Question

In one experiment, the reaction of \(1.00 \mathrm{g}\) mercury and an excess of sulfur yielded \(1.16 \mathrm{g}\) of a sulfide of mercury as the sole product. In a second experiment, the same sulfide was produced in the reaction of \(1.50 \mathrm{g}\) mercury and \(1.00 \mathrm{g}\) sulfur. (a) What mass of the sulfide of mercury was produced in the second experiment? (b) What mass of which element (mercury or sulfur) remained unreacted in the second experiment?

Short answer

The mass of the sulfide of mercury produced in the second experiment was 1.74g and the mass of sulfur that remained unreacted was 0.76g.

Step-by-step solution

Calculate Mass of Sulfide Produced in Second Experiment

The amount of mercury is 1.5 g in the second experiment. As the sulfide of mercury is produced by the reaction of 1g of mercury, we can calculate the mass of the sulfide produced in the second experiment by setting up a proportion: \(1.00 g \, mercury = 1.16 g \, sulfide \) and \(1.50 g \, mercury = x g \, sulfide\), thus solving for x gives \(x = 1.5 * \frac{1.16}{1} = 1.74 g \,of\, sulfide\).

Calculate Mass of Reacted Sulfur

From the first experiment, we know that the mass of sulfur which reacted with 1g of mercury is \(1.16 g - 1 g = 0.16 g\). Thus, the mass of the sulfur reacted with 1.5g of mercury in the second experiment can again be calculated using a proportion. This gives: \(1 g \, mercury = 0.16 g \, sulfur\) and \(1.5 g \, mercury= y g\, sulfur \), thus solving for y gives \(y = 1.5 * \frac{0.16}{1} = 0.24 g \,of \, reacted \,sulfur\).

Determine Element Remaining Unreacted

In the second experiment, we started with 1 g of sulfur. The mass of reacted sulfur was calculated earlier to be 0.24 g, so the amount of sulfur remaining is \(1 g - 0.24 g = 0.76 g\). As there is still sulfur left, but no mercury, sulfur is the element that remained unreacted.

Key concepts

Understanding Reaction Yield

In chemical reactions, the reaction yield represents the amount of product formed through the interaction of reactants under specific conditions. It is a crucial concept in stoichiometry, as it tells us how efficiently the reactants have converted into products. In the given experiment, the formation of sulfide of mercury can help us calculate the yield of the reaction.
Using the example, we see that 1.00 g of mercury reacted with sulfur to yield 1.16 g of the mercury sulfide product. This sets the scene for understanding how changes in reactant quantities affect the yield in the second experiment. By proportionately scaling up the amount of mercury, we observe from the solution that using 1.50 g of mercury would result in a product yield of 1.74 g of mercury sulfide.
Calculating reaction yield helps in optimizing chemical processes to achieve maximum efficiency. Imagine scaling this reaction to an industrial level where every gram of product counts. Thus, knowing the potential yield allows chemists to better plan and control reactions for desired outcomes.

Determining the Limiting Reactant

In stoichiometry, the limiting reactant is the substance that is entirely consumed first, thereby determining the maximum amount of product that can be formed. In our exercise, it's crucial to identify which reactant limits the reaction to predict the unreacted material.
In the first experiment, sulfur was in excess, allowing full conversion of mercury to mercury sulfide. However, in the second experiment, where mercury was increased, identifying the limiting reactant becomes essential.
We started with 1.50 g of mercury and 1.00 g of sulfur. By considering the stoichiometry established in the first reaction (1 g mercury reacts with 0.16 g sulfur), we calculated 0.24 g of sulfur would react with 1.50 g of mercury. Since only 0.24 g of sulfur could react, leaving 0.76 g unreacted, sulfur remained the excess reactant, making mercury the limiting reactant in the second experiment.
This understanding is vital in chemical reactions, especially when reactants are costly or hard to obtain. It ensures maximum productivity and cost-effectiveness, guiding chemists in adjusting the amounts used in reactions.

Principles of Mass Conservation

The mass conservation principle is a fundamental concept in chemical reactions, stating that mass cannot be created or destroyed in an isolated system. This principle is instrumental in verifying the accuracy of stoichiometric calculations.
In the context of the given exercise, the total mass before and after the reaction must be equal, which is evidenced by the stoichiometric calculations performed. Initially, in the second experiment, we have a total of 2.50 g of reactants (1.50 g of mercury plus 1.00 g of sulfur). The calculations indicate this should lead to 1.74 g of mercury sulfide, aligning with the principle of mass conservation.
Additionally, understanding that the unreacted sulfur is 0.76 g supports this principle, as 1.74 g (product) + 0.76 g (unreacted sulfur) = 2.50 g (initial total mass). Students can correlate this with Balancing Chemical Equations, where the total mass of reactants equals the mass of products plus leftover reactants.

• Helps verify the correctness of chemical equations and calculations.
• Critical in predicting reactions in closed systems.
• Ensures that all mass is accounted for, crucial in lab settings to avoid material loss.